x+y=−2 xy=4 find x^8 +8y^5 =?


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Question Number 169865 by mr W last updated on 11/May/22

$x + y = - 2$ $x y = 4$ $f i n d x^{8} + 8 y^{5} = ?$
Commented by cortano1 last updated on 11/May/22
$y=−x−2 ∧ x(−x−2)=4 ⇒x^2 +2x+4=0 ⇒(x+1)^2 +3 = 0 ⇒x=−1± i(√3) ⇒x=2 (−(1/2) ± ((√3)/2) i) ⇒x=−2((1/2) ∓ ((√3)/2) i)=−2e^(± i(π/3)) ⇒x^8 = 256 (cos ((8π)/3)+i sin ((8π)/3)) ⇒x^8 = 256 (cos ((2π)/3) + i sin ((2π)/3)) ⇒x^8 = 256 (−(1/2)+(1/2)(√3) i)=−128+128(√3) i ⇒y=−x−2=1∓ i(√3) −2=−1∓ i(√3) ⇒y=−2((1/2)±((√3)/2) i)=−2e^(±i (π/3)) ⇒y^5 = −32(cos ((5π)/3) ± i sin ((5π)/3)) ⇒y^5 =−32(−cos ((2π)/3) ∓ i sin ((2π)/3)) ⇒y^5 = −32((1/2) ∓ ((√3)/2) i)=−16 ± 16(√3) i ⇒8y^5 = −128 ±128(√3) i { ((x^8 +8y^5 = −256+256(√3) i)),((x^8 +8y^5 = −256)) :}$
$y = - x - 2 \land x (- x - 2) = 4$ $\Rightarrow x^{2} + 2 x + 4 = 0$ $\Rightarrow {(x + 1)}^{2} + 3 = 0$ $\Rightarrow x = - 1 \pm i \sqrt{3}$ $\Rightarrow x = 2 (- \frac{1}{2} \pm \frac{\sqrt{3}}{2} i)$ $\Rightarrow x = - 2 (\frac{1}{2} \mp \frac{\sqrt{3}}{2} i) = - 2 e^{\pm i \frac{π}{3}}$ $\Rightarrow x^{8} = 256 (\cos \frac{8 π}{3} + i \sin \frac{8 π}{3})$ $\Rightarrow x^{8} = 256 (\cos \frac{2 π}{3} + i \sin \frac{2 π}{3})$ $\Rightarrow x^{8} = 256 (- \frac{1}{2} + \frac{1}{2} \sqrt{3} i) = - 128 + 128 \sqrt{3} i$ $\Rightarrow y = - x - 2 = 1 \mp i \sqrt{3} - 2 = - 1 \mp i \sqrt{3}$ $\Rightarrow y = - 2 (\frac{1}{2} \pm \frac{\sqrt{3}}{2} i) = - 2 e^{\pm i \frac{π}{3}}$ $\Rightarrow y^{5} = - 32 (\cos \frac{5 π}{3} \pm i \sin \frac{5 π}{3})$ $\Rightarrow y^{5} = - 32 (- \cos \frac{2 π}{3} \mp i \sin \frac{2 π}{3})$ $\Rightarrow y^{5} = - 32 (\frac{1}{2} \mp \frac{\sqrt{3}}{2} i) = - 16 \pm 16 \sqrt{3} i$ $\Rightarrow 8 y^{5} = - 128 \pm 128 \sqrt{3} i$ ${\begin{cases} x^{8} + 8 y^{5} = - 256 + 256 \sqrt{3} i \\ x^{8} + 8 y^{5} = - 256 \end{cases}$
Commented by infinityaction last updated on 13/May/22
$2x^8 + 8(2y^5 ) = 2p(let) (x^8 + y^8 + x^8 − y^8 ) _(A) + 8(y^5 + x^5 + y^5 − x^5 )_(B) = 2p 2p = A + 8B A = x^8 +y^(8 ) +(x^4 +y^4 )(x^2 +y^2 )(x+y)(x−y) ∴ x+y = −2 and xy = 4 x^2 + y^2 = 4−2xy = −4 x^(4 ) + y^4 = −16 x^8 + y^8 = −256 A = −256 −128(x−y) B = y^5 +x^5 +y^5 −x^5 B=(x^4 +y^4 )(x+y)−xy^4 −yx^4 +(x^4 +y^4 )(y−x)−yx^4 +xy^4 B= 32−4(x^3 +y^3 )−16(y−x)−xy(x^3 −y^3 ) B = 32−4(x+y)(x^2 +y^2 −xy)+16(x−y)+4(y−x)(y^2 +x^2 +xy) B= 32+4(−2)×8+16(x−y)+4(y−x)(−4+4) B = −32 + 16(x−y) 2p = −256−128(x−y)+8{−32 +16(x−y)} 2p = −256−128(x−y)−256+128(x−y) 2p = −512 p = −256$
$2 x^{8} + 8 (2 y^{5}) = 2 p (l e t)$ $\underset{A}{\underset{⏟}{(x^{8} + y^{8} + x^{8} - y^{8})}} + 8 \underset{B}{\underset{⏟}{(y^{5} + x^{5} + y^{5} - x^{5})}} = 2 p$ $2 p = A + 8 B$ $A = x^{8} + y^{8} + (x^{4} + y^{4}) (x^{2} + y^{2}) (x + y) (x - y)$ $∴ x + y = - 2 a n d x y = 4$ $x^{2} + y^{2} = 4 - 2 x y = - 4$ $x^{4} + y^{4} = - 16$ $x^{8} + y^{8} = - 256$ $A = - 256 - 128 (x - y)$ $B = y^{5} + x^{5} + y^{5} - x^{5}$ $B = (x^{4} + y^{4}) (x + y) - {x y}^{4} - {y x}^{4} + (x^{4} + y^{4}) (y - x) - {y x}^{4} + {x y}^{4}$ $B = 32 - 4 (x^{3} + y^{3}) - 16 (y - x) - x y (x^{3} - y^{3})$ $B = 32 - 4 (x + y) (x^{2} + y^{2} - xy) + 16 (x - y) + 4 (y - x) (y^{2} + x^{2} + xy)$ $B = 32 + 4 (- 2) \times 8 + 16 (x - y) + 4 (y - x) (- 4 + 4)$ $B = - 32 + 16 (x - y)$ $2 p = - 256 - 128 (x - y) + 8 {- 32 + 16 (x - y)}$ $2 p = - 256 - 128 (x - y) - 256 + 128 (x - y)$ $2 p = - 512$ $p = - 256$
	Answered by Rasheed.Sindhi last updated on 11/May/22
	${: ((x+y=−2)),((xy=4)) }x=−1±i(√3) x(−x−2)=4 x^2 +2x+4=0,( Similarly y^2 +2y+4=0) x^2 =−2x−4 , (y^2 =−2y−4) •x^8 =(x^2 )^4 =(−2x−4)^4 =2^4 (x+2)^4 =16(x^2 +4x+4)^2 =16(−2x−4+4x+4)^2 = =16(2x)^2 =64(−2x−4)=−128(x+2) =−128(−1±i(√3)+2)=−128(1±i(√3) ) •8y^5 = 8(y^2 )^2 y=8y(−2y−4)^2 =16y(y^2 +4y+4)=16y(−2y−4+4y+4) =16y(2y) =32y^2 =32(−2y−4)=−64(y+2) =64(−x−2+2)=−64x=−64(−1±i(√3)) •x^8 +8y^5 =−128(1±i(√3) )+−64(−1±i(√3)) =−128+64∓128i(√3) ∓64i(√3) =−64∓192i(√3) ??$
	$\begin{matrix} x + y = - 2 \\ x y = 4 \end{matrix}} x = - 1 \pm i \sqrt{3}$ $x (- x - 2) = 4$ $x^{2} + 2 x + 4 = 0, (S i m i l a r l y y^{2} + 2 y + 4 = 0)$ $x^{2} = - 2 x - 4, (y^{2} = - 2 y - 4)$ $∙ x^{8} = {(x^{2})}^{4} = {(- 2 x - 4)}^{4} = 2^{4} {(x + 2)}^{4}$ $= 16 {(x^{2} + 4 x + 4)}^{2} = 16 {(- 2 x - 4 + 4 x + 4)}^{2} =$ $= 16 {(2 x)}^{2} = 64 (- 2 x - 4) = - 128 (x + 2)$ $= - 128 (- 1 \pm i \sqrt{3} + 2) = - 128 (1 \pm i \sqrt{3})$ $∙ 8 y^{5} = 8 {(y^{2})}^{2} y = 8 y {(- 2 y - 4)}^{2}$ $= 16 y (y^{2} + 4 y + 4) = 16 y (- 2 y - 4 + 4 y + 4)$ $= 16 y (2 y)$ $= 32 y^{2} = 32 (- 2 y - 4) = - 64 (y + 2)$ $= 64 (- x - 2 + 2) = - 64 x = - 64 (- 1 \pm i \sqrt{3})$ $∙ x^{8} + 8 y^{5}$ $= - 128 (1 \pm i \sqrt{3}) + - 64 (- 1 \pm i \sqrt{3})$ $= - 128 + 64 \mp 128 i \sqrt{3} \mp 64 i \sqrt{3}$ $= - 64 \mp 192 i \sqrt{3} ? ?$
	Commented by greougoury555 last updated on 11/May/22

	$16 {(x^{2} - 4 x + 4)}^{2} = 16 {(- 6 x)}^{2} = 16 \times 36 x^{2}$
	Commented by Rasheed.Sindhi last updated on 11/May/22

	$T h a n k s s i r f o r p o i n t i n g o u t m y m i s t a k e .$
	Answered by Rasheed.Sindhi last updated on 13/May/22
	${ ((x+y=−2)),((xy=4)) :} ;x^8 +8y^5 =? { ((x+y=−2⇒x=−y−2)),((xy=4⇒y(−y−2)=4=y^2 +2y+4=0⇒y^2 =−2y−4)) :} x^8 +8y^5 =(−y−2)^8 +8y^5 =(y^2 +4y+4)^4 +8y(y^2 )^2 =(−2y−4+4y+4)^4 +8y(−2y−4)^2 =(2y)^4 +32y(y+2)^2 =16(y^2 )^2 +32y(y^2 +4y+4) =16(−2y−4)^2 +32y(−2y−4+4y+4) =64(y^2 +4y+4)+32y(2y) =64(−2y−4+4y+4)+64y^2 =64(2y)+64(−2y−4) =128y−128y−256 =−256$
	${\begin{cases} x + y = - 2 \\ x y = 4 \end{cases}; x^{8} + 8 y^{5} = ?$ ${\begin{cases} x + y = - 2 \Rightarrow x = - y - 2 \\ x y = 4 \Rightarrow y (- y - 2) = 4 = y^{2} + 2 y + 4 = 0 \Rightarrow y^{2} = - 2 y - 4 \end{cases}$ $x^{8} + 8 y^{5} = {(- y - 2)}^{8} + 8 y^{5}$ $= {(y^{2} + 4 y + 4)}^{4} + 8 y {(y^{2})}^{2}$ $= {(- 2 y - 4 + 4 y + 4)}^{4} + 8 y {(- 2 y - 4)}^{2}$ $= {(2 y)}^{4} + 32 y {(y + 2)}^{2}$ $= 16 {(y^{2})}^{2} + 32 y (y^{2} + 4 y + 4)$ $= 16 {(- 2 y - 4)}^{2} + 32 y (- 2 y - 4 + 4 y + 4)$ $= 64 (y^{2} + 4 y + 4) + 32 y (2 y)$ $= 64 (- 2 y - 4 + 4 y + 4) + 64 y^{2}$ $= 64 (2 y) + 64 (- 2 y - 4)$ $= 128 y - 128 y - 256$ $= - 256$
	Answered by mr W last updated on 13/May/22

	$x, y a r e r o o t s o f z^{2} + 2 z + 4 = 0 .$ $z^{2} = - 2 z - 4$ $z^{3} = - 2 z^{2} - 4 z = - 2 (- 2 z - 4) - 4 z = 8$ $i . e . x^{3} = y^{3} = 8$ $x^{8} + 8 y^{5}$ $= x^{3} x^{3} x^{2} + 8 y^{3} y^{2}$ $= 8 \times 8 x^{2} + 8 \times 8 y^{2}$ $= 64 (x^{2} + y^{2})$ $= 64 [{(x + y)}^{2} - 2 x y]$ $= 64 [{(- 2)}^{2} - 2 \times 4]$ $= - 64 \times 4$ $= - 256 ✓$
	Commented by infinityaction last updated on 13/May/22

	$g r e a t s i r v e r y n i c e$
	Commented by peter frank last updated on 13/May/22

	$short and clear$
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